Reading C Declarations: A Guide for the Mystified

By Eric Giguere
October 1987

Introduction

One of the most confusing aspects of the C programming language
is the way in which variable types are declared. These type
declarations manage to mystify even experienced programmers. This
guide is intended to help in deciphering those declarations,
including the new twists introduced by the ANSI Standard. Several
examples are given to illustrate the methods, and a short quiz at the
end allows you to test yourself on a variety of sample declarations.

What is a declaration?

Throughout this guide the word "declaration" will be used to
refer to a variable declaration statement. The following two
statements are both declarations:

static int x;
register short y = 0;

Pre-ANSI Declarations

A C type declaration can be broken into two distinct parts: a
type specifier and a declarator. The type specifier is the easy
part: it declares the (optional) storage class (either auto,
static, register or extern) and a type (such
as int or unsigned char). The
declarator is the part that causes confusion, and specifies whether a
variable is to be considered a simple type, a pointer, an array, a
function, or a combination of these. The identifier, the variable
name, is found in the middle of the declarator.

The rules we'll use will allow you to follow a declaration by
writing a phrase on a piece of paper at each step in the decoding
process. The final result will be a phrase "declare variable-name as
..." stating in plain English what exactly is being declared.

Rules

Find the identifier. This is your starting point. On a piece
of paper, write "declare identifier as".

Look to the right. If there is nothing there, or there is a
right parenthesis ")", goto step 4.

You are now positioned either on an array (left bracket) or
function (left parenthesis) descriptor. There may be
a sequence of these, ending either with an unmatched right
parenthesis or the end of the declarator (a semicolon or
a "=" for initialization). For each such descriptor,
reading from left to right:

if an empty array "[]", write "array of"

if an array with a size, write "array size of"

if a function "()", write "function returning"

Stop at the unmatched parenthesis or the end of the declarator,
whichever comes first.

Return to the starting position and look to the left. If there
is nothing there, or there is a left parenthesis "(", goto step 6.

You are now positioned on a pointer descriptor, "*". There may
be a sequence of these to the left, ending either with an
unmatched left parenthesis "(" or the start of the declarator.
Reading from right to left, for each pointer descriptor
write "pointer to". Stop at the unmatched parenthesis or the
start of the declarator, whichever is first.

At this point you have either a parenthesized expression or the
complete declarator. If you have a parenthesized expression,
consider it as your new starting point and return to step 2.

Write down the type specifier. Stop.

Examples

The rules may seem confusing, so let's apply them to a few
examples:

Example 1:static int *x;

Our starting point is the identifier x. We write "declare x as". We
look to the right. Nothing there, goto step 4. We have a pointer
descriptor, so we goto step 5. We write "pointer to" and stop since
we've reached the start of the declarator. At step 6 we see we have
the complete declarator, so we move to step 7 and write "static int"
and stop.

The declaration is: "declare ptar as pointer to array 10 of int".
Here the parentheses were used to force the pointer descriptor to
have precendence over the array descriptor. The array has ten
elements.

The declaration is: "declare fp as pointer to function returning
int". Function and array descriptors have the same precendence, so
again the parentheses were necessary to force the pointer descriptor
to be read first. (Note then that int *fp(); declares fp as a
function returning pointer to int.)

The declaration is: "declare list as array MAX of pointer to function
returning pointer to int". Without the parentheses we would have had
an array of functions returning pointers to pointers to int, which is
illegal in C because arrays cannot hold functions.

The declaration is "declare table as array 10 of array 20 of pointer
to char".

The important thing to remember while reading these declarations
is that functions and arrays are read left-to-right and have
precedence over pointers, which are read right-to-left.

ANSI Additions

The ANSI Standard for C makes a few changes to what a C type
declaration can look like. The basic structure of type specifier and
declarator remains the same. However, a function designator may now
include both the number and the types of parameters the function
accepts and pointers may also be modified by the keywords const and
volatile.

A further addition is the abstract declarator, used to describe
function parameters. An abstract declarator is a declarator minus
the identifier. Thus instead of *a[] you could
conceivably use *[].
This makes it a bit harder to spot the correct starting position in
the declaration. Note that abstract declarators can only be used in
certain situations  an identifier is a necessary part of most
declarations.

The rules given in the previous section need to be amended
slightly to handle these new additions, but rather than list them
we'll go through a few examples.

Example 7:int func( char *, int );

Step 1, find the identifier, write "declare func as". Step 2, notice
the function descriptor to the right, goto step 3. Here we make some
changes. Write simply "function". Now shift your attention inside
the parentheses making up the function descriptor. These are the
types of parameters that the function expects. Each parameter is a
declaration in itself, with a type specifier and an abstract
declarator. Put aside the current declaration for now, write "(with
first parameter of type", and goto step 2 with char * as the new
starting point:

There are no more parameters to be dealt with so we continue with our
original declaration, first writing ") returning" and continue where
we left off in step 3. Step 4, no pointers on the left. Step 6, end
of declaration. Step 7, write "int". Stop.

The complete declaration is "declare func as function (with first
parameter of type pointer to char and with second parameter of type
int) returning int."

You can (and probably should) ignore the parameters if you wish. This
would then give us "declare func as function returning int". All we
are doing when declaring the parameters is adding more detail.

Example 8:const int *ptr1;

This case is handled using the old rules to give us "declare ptr1 as
a pointer to const int", where const means the the int cannot be
modified.

Example 9:int *const ptr2;

This is an example of the new syntax involving const and volatile.
Either or both of these two types may be placed after a pointer
descriptor to assign those attributes to the pointer itself. Back to
the rules: Step 1, start at ptr2, the only choice for identifier
since const and volatile are reserved words in ANSI C, write "declare
ptr2 as". Step 2, nothing to the right, goto step 4. Step 4, we
have a "*" followed by const. This qualifies the pointer descriptor.
Step 5, write "const pointer to". Step 6, end of declaration. Step
7, write "int". Stop.

The declaration is: "declare ptr2 as const pointer to int". The
distinction to make between this declaration and the one in Example 8
is that the const attribute is associated with the pointer, and not
the type specifier. So while the pointer itself cannot be modified,
the type it points to can.

The volatile type can also be used in this fashion, either by itself
or in combination with const. Write it the same way.

Pointer descriptors do not require either of these attributes,
however, so a declaration of the form int **a; is still quite valid.

Typing Overload

Obviously, the more information is added to a declaration,
the harder it gets to read.

Example 10:extern char *const (*goop( char *b ))( int, long );

Where do we start? There are two identifiers in this declaration,
goop and b. Notice, however, that char *b is declaring a function
parameter; goop is our starting point. Step 1, write "declare goop
as". Step 2, function on the right. Step 3, write "function". Now
look inside the function to find any parameters. Write "(with one
parameter of type char *". Notice how we ignore the identifier b.
Function parameters declared in function prototypes can be declared
with abstract or normal declarators; in the latter case the
identifier is ignored. Only the typing information counts. (The
identifiers are not ignored in ANSI function definitions, but we're
declaring a prototype here.)

Now go back to step 3, complete the function and write ") returning".
Step 4, a pointer to the left of goop. Step 5, write "pointer to".
Step 6, a parenthesized expression, back to step 2. Step 2, function
on the right. Step 3, write "function". Now go inside, write "(with
first parameter of type int and with second parameter of type long)".
End of function, back to step 3 and write "returning". Step 4,
pointer (with const) on the left. Step 5, write "const pointer to".
Step 6, end of declaration. Step 7, write "extern char". Stop.

The declaration is (take a deep breath): "declare goop as function
(with one parameter of type char *) returning pointer to function
(with first parameter of type int and second parameter of type long)
returning const pointer to extern char." (Whew!)

Or if you choose to ignore the parameters: "declare goop as function
returning pointer to function returning const pointer to extern
char", which is a bit less intimidating.

Parentheses

One of the harder things to do is to decide what a given set of
parentheses imply  are we declaring a function or are we simply
grouping an expression? The simplest way to differentiate the two is
to look at the placement of the type specifiers. If a type specifier
immediately follows a left parenthesis, that parenthesis is the start
of a function descriptor and the type is part of a function
parameter. Empty parentheses also signify a function. Otherwise the
parentheses are grouping (perhaps unnecessarily, but it's better to
have too many than too few) the expression.

The Quiz

After all this work, have we accomplished anything? You might
think that working out those examples step-by-step was overkill.
Maybe. But read them over carefully once or twice, work through the
following examples, and you won't ever be confused by a C declaration
again. Answers can be found below.

char ****q[ 30 ];

char **(**q)[ 30 ];

extern int (x)[];

long (*a[])( char, char );

int *(*(*(*b)())[10])();

char *strprt( char (*)( int ), unsigned char );

int (*const ab[])( unsigned int );

Answers

declare q as array 30 of pointer to pointer to pointer to
pointer to char

declare q as pointer to pointer to array 30 of pointer to
pointer to char

declare x as array of extern int

declare a as array of pointer to function (with two parameters
of type char) returning long

declare b as pointer to function returning pointer to array 10 of
pointer to function returning pointer to int

declare strprt as function (with first parameter of type pointer
to function (with parameter of type int) returning char and with
second parameter of type unsigned char) returning pointer to char